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Algebra Level 4

6 + 2 3 33 − 19 3 \large \frac{\sqrt{6+2\sqrt3}}{\sqrt{33-19\sqrt3}}

If the value of the radical expression above equals to a + b 3 a+ b \sqrt 3 for integers a a and b b , find the value of a + b a+b .

6 None of these choices 12 10 8

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Harshi Singh by Harshi Singh , 20, India

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2 solutions

Ikkyu San
Apr 30, 2015

6 + 2 3 33 − 19 3 = 6 + 2 3 33 − 19 3 = ( 6 + 2 3 ) ( 33 + 19 3 ) ( 33 − 19 3 ) ( 33 + 19 3 ) = 198 + 114 3 + 66 3 + 38 ( 3 ) 3 3 2 − ( 19 3 ) 2 = 198 + 114 + 180 3 1089 − 1083 = 312 + 180 3 6 = 52 + 30 3 = 25 + 27 + 2 ( 5 ) ( 3 3 ) = 25 + 2 ( 5 ) ( 3 3 ) + 27 = 5 2 + 2 ( 5 ) ( 3 3 ) + ( 3 3 ) 2 = ( 5 + 3 3 ) 2 = 5 + 3 3 \begin{aligned}\dfrac{\sqrt{6+2\sqrt{3}}}{\sqrt{33-19\sqrt{3}}}=&\sqrt{\dfrac{6+2\sqrt{3}}{33-19\sqrt{3}}}\\=&\sqrt{\dfrac{(6+2\sqrt{3})\color{#624F41}{(33+19\sqrt{3})}}{(33-19\sqrt{3})\color{#624F41}{(33+19\sqrt{3})}}}\\=&\sqrt{\dfrac{198+114\sqrt{3}+66\sqrt{3}+38(3)}{33^2-(19\sqrt{3})^2}}\\=&\sqrt{\dfrac{198+114+180\sqrt{3}}{1089-1083}} \\=&\sqrt{\dfrac{312+180\sqrt{3}}{6}} \\=&\sqrt{\color{#D61F06}{52}+\color{#3D99F6}{30\sqrt{3}}} \\=&\sqrt{\color{#D61F06}{25+27}+\color{#3D99F6}{2(5)(3\sqrt{3})}}\\=&\sqrt{\color{#20A900}{25}+2(5)(3\sqrt{3})+\color{magenta}{27}} \\=&\sqrt{\color{#20A900}{5^2}+2(\color{#20A900}{5})(\color{magenta}{3\sqrt{3}})+\color{magenta}{(3\sqrt{3})^2}}\\ =&\sqrt{(\color{#20A900}{5}+\color{magenta}{3\sqrt{3}})^2}\\ =&\ \color{#20A900}{5}+\color{magenta}{3}\sqrt{3} \end{aligned}

Therefore, a = 5 , b = 3 , a + b = 5 + 3 = 8 a=\color{#20A900}{5}, b=\color{magenta}{3}, a+b=\color{#20A900}{5}+\color{magenta}{3}=\boxed{8}

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nice answer firstly i didnt get this question but now it seems easy .thanks

Harshi Singh - 6 years, 1 month ago

Did the same way....nyc solution

Apoorv Singhal - 6 years ago

6 + 2 3 33 − 19 3 = 2 3 + 2 11 3 − 19 = ( 2 3 + 2 ) ( 11 3 + 19 ) ( 11 3 − 19 ) ( 11 3 + 19 ) = 52 + 30 3 = a + b 3 ∴ 52 + 30 3 = a 2 + 3 b 2 + 2 a b 3 . ⟹ a b = 15 = 5 ∗ 3 , a n d a 2 + 3 b 2 = 52 S o a = 5 , a n d b = 3. a + b = 8. \dfrac{\sqrt{6+2\sqrt{3}}}{\sqrt{33-19\sqrt{3}}}\\ =\dfrac{\sqrt{2\sqrt{3}+2}}{\sqrt{11\sqrt{3} -19}} \\ =\sqrt{ \dfrac {(2\sqrt{3}+2)(11\sqrt{3} +19) } {(11\sqrt{3} -19)(11\sqrt{3} +19) } } \\ =\sqrt{ 52+30\sqrt{3} }~= ~a+b\sqrt{3} \\ \therefore~~52+30\sqrt3=a^2+3b^2+2ab\sqrt3.\\ \implies~~ ab=15=5*3,~~~~~and~~~~a^2+3b^2=52\\ So~a=5,~~and~~b=3.\\ a+b=\color{#D61F06}{8}.\\

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